Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{q^2 + 6q + 8}{-2q^2 - 2q + 24} \div \dfrac{q + 2}{5q + 20} $
Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{q^2 + 6q + 8}{-2q^2 - 2q + 24} \times \dfrac{5q + 20}{q + 2} $ First factor out any common factors. $a = \dfrac{q^2 + 6q + 8}{-2(q^2 + q - 12)} \times \dfrac{5(q + 4)}{q + 2} $ Then factor the quadratic expressions. $a = \dfrac {(q + 4)(q + 2)} {-2(q + 4)(q - 3)} \times \dfrac {5(q + 4)} {q + 2} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac { (q + 4)(q + 2) \times 5(q + 4)} { -2(q + 4)(q - 3) \times (q + 2)} $ $a = \dfrac {5(q + 4)(q + 2)(q + 4)} {-2(q + 4)(q - 3)(q + 2)} $ Notice that $(q + 4)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {5\cancel{(q + 4)}(q + 2)(q + 4)} {-2\cancel{(q + 4)}(q - 3)(q + 2)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $a = \dfrac {5\cancel{(q + 4)}\cancel{(q + 2)}(q + 4)} {-2\cancel{(q + 4)}(q - 3)\cancel{(q + 2)}} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac {5(q + 4)} {-2(q - 3)} $ $ a = \dfrac{-5(q + 4)}{2(q - 3)}; q \neq -4; q \neq -2 $